is the minimum polynomial, it has degree at least as large
as n. However, p
(λ)
divides q
(λ)
because q
(C)
= 0.
p(λ) = q(λ)l(λ)+ r(λ)
where r
(λ)
= 0 or it has degree less than n = deg
(q(λ))
. The above equation shows that
r
(C )
= 0 and this would contradict p
(λ)
being the minimum polynomial. Thus q
(λ)
∕p
(λ)
.
Also,
q(λ) = p(λ)l(λ)+ ˆr(λ)
where
ˆr
(λ)
has smaller degree than p
(λ)
or else is 0. Again, if it is nonzero, you would have a contradiction
to p
(λ)
being the minimum polynomial. Thus
ˆr
(λ)
= 0 and so p
(λ)
∕q
(λ )
. These are both monic
polynomials and so they must be the same. ■
Now recall the rational canonical form. You have A ∈ℒ
(V,V)
where V is a finite dimensional vector
space. The minimum polynomial is
r∏ ki
p (λ) = ϕi (λ)
i=1
where ϕ_{i}
(λ)
is a monic irreducible polynomial of degree d. Then for V_{i}≡ ker
( ki)
ϕi(A)
, you
have
r
V =i=1 Vi
where each V_{i} is an A invariant subspace of V and the minimum polynomial of A restricted to V_{i} is
ϕ_{i}
(λ)
^{ki}. Then to get the rational canonical form, you consider a cyclic basis for each V_{i} and obtain
this as a block diagonal matrix of block diagonal matrices. The large blocks pertain to the
individual V_{i} and the question of uniqueness reduces to consideration of the restriction of A on
V_{i}. Thus the problem reduces to A ∈ℒ
(V,V)
and its minimum polynomial is ϕ
(λ)
^{k} where
ϕ
(λ )
is monic and irreducible. Do any two rational canonical forms for this situation have the
same blocks? Recall that the blocks are obtained from cycles of length md which are of the
form
( md−1 )
x Ax ⋅⋅⋅ A x
where A^{md}x = −
( )
a0 + a1Ax + ⋅⋅⋅+ am −1Amd−1x
where ϕ
(λ )
^{m} = a_{0} + a_{1}λ +
⋅⋅⋅
+ a_{m−1}λ^{md−1} + λ^{md}.
Then the block obtained is C
(ϕ(λ)m)
obtained as follows:
( ) ( )
Ax A2x ⋅⋅⋅ Amdx = x Ax ⋅⋅⋅ Amd− 1x C (ϕ (λ)m )
Suppose then that you have two rational canonical forms from the same A.
( )
C (ϕ (λ)m1 ) 0
|| C (ϕ(λ)m2) ||
M ≡ || . || ,
( .. )
0 C (ϕ (λ)mr )
( n1 )
| C (ϕ (λ) ) n2 0 |
|| C (ϕ (λ) ) ||
N ≡ |( ... |)
ns
0 C (ϕ(λ) )
where in each case, the size of the blocks are decreasing from upper left to lower right. Why is r = s and
m_{i} = n_{i}? Since these come from the same linear transformation with possibly different bases, this is the
same as saying that they are similar. Furthermore, if f
(λ)
is a polynomial, then f
(M )
must also be similar
to f
(N )
. By block multiplication this would say that
( m1 )
| f (C (ϕ(λ) )) m 0 |
|| f (C(ϕ (λ) 2)) ||
f (M ) ≡ |( ... |) ,
mr
0 f (C (ϕ(λ) ))
( f (C (ϕ(λ)n1)) 0 )
|| f (C (ϕ(λ)n2)) ||
f (N) ≡ || . ||
( .. )
0 f (C (ϕ(λ)ns))
are similar. Now we pick f auspiciously. Suppose the blocks are the same size for the first k − 1 blocks
starting at the top and going toward the bottom, and that there is a discrepancy in the k^{th} position. Then
the first k − 1 blocks must be identical before the discrepancy. Say C
These cannot possibly be similar since they do not even have the same rank and so there can be no
discrepancies in the size of the blocks. Hence they are all identical. This proves the following
theorem.
Theorem 7.5.3Let A ∈ℒ
(V,V )
. Then its rationalcanonical form is uniquely determined up tothe order of the blocks. In particular, the numbers
|βx|
occuring in the cyclic basis are determined.
In the case where two n × n matrices M,N are similar, recall this is equivalent to the two being
matrices of the same linear transformation taken with respect to two different bases. Hence each are similar
to the same rational canonical form.
Example 7.5.4Here is a matrix.
( )
5 − 2 1
A = |( 2 10 − 2 |)
9 0 9
Find a similarity transformation which will produce the rational canonical form for A.
The minimum polynomial is λ^{3}− 24λ^{2} + 180λ − 432. This factors as
(λ− 6)2(λ− 12)
Thus ℚ^{3} is the direct sum of ker
( )
(A − 6I)2
and ker
(A − 12I)
. Consider the first of these. You see easily
that this is
What about the length of A cyclic sets? It turns out it doesn’t matter much. You can start
with either of these and get a cycle of length 2. Lets pick the second one. This leads to the
cycle
where the last of the three is a linear combination of the first two. Take the first two as the first two
columns of S. To get the third, you need a cycle of length 1 corresponding to ker
(A− 12I)
. This yields the
eigenvector
( )
1 − 2 3
^{T}. Thus
( )
| − 1 − 4 1 |
S = ( 0 − 4 − 2 )
1 0 3
Now using Proposition 5.2.9, the rational canonical form for A should be
^{2} = λ^{2}− 12λ + 36 and so the top left block is indeed the companion matrix of this
polynomial. Of course in this case, we could have obtained the Jordan canonical form.
because the next vector involving A^{2} yields a vector which is in the span of the above two. You
check this by making the vectors the columns of a matrix and finding the row reduced echelon
form. Clearly this cycle does not span ker
(ϕ2(A))
, so look for another cycle. Begin with a
vector which is not in the span of these two. The last one works well. Thus another A cycle
is
and you see this is in rational canonical form, the two 2 × 2 blocks being companion matrices for the
polynomial λ^{2}− 8λ + 32 and the 1 × 1 block being a companion matrix for λ− 4. Note that you could have
written this without finding a similarity transformation to produce it. This follows from the above theory
which gave the existence of the rational canonical form.
Obviously there is a lot more which could be considered about rational canonical forms. Just
begin with a strange field and start investigating what can be said. One can also derive more
systematic methods for finding the rational canonical form. The advantage of this is you don’t need
to find the eigenvalues in order to compute the rational canonical form and it can often be
computed for this reason, unlike the Jordan form. The uniqueness of this rational canonical
form can be used to determine whether two matrices consisting of entries in some field are
similar, provided you are able to factor the minimum polynomial into the product of irreducible
factors.